All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
A repunit number (Posted on 2011-12-26) Difficulty: 2 of 5
3 integers form an arithmetic progression with d=2 i.e. a, a+2, a+4.
The sum of their squares is a 4-digit number divisible by 1111.
List all possible triplets.

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): is there a mistake in the puzzle? Comment 5 of 5 |
(In reply to re: is there a mistake in the puzzle? by Jer)

I goofed.
  Posted by Charlie on 2011-12-26 14:54:09

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (25)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information